动态规划之复制粘贴数

LeetCode 650. 2 Keys Keyboard

题目

Initially on a notepad only one character 'A' is present. You can perform two operations on this notepad for each step:

Copy All

: You can copy all the characters present on the notepad (partial copy is not allowed).

Paste

: You can paste the characters which are copied last time.

Given a number 

n

. You have to get exactly 

n

 'A'
on the notepad by performing the minimum number of steps permitted. Output the minimum number of steps to get 

n

 'A'.

Example 1:

Input: 3
Output: 3
Explanation:
Intitally, we have one character 'A'.
In step 1, we use Copy All operation.
In step 2, we use Paste operation to get 'AA'.
In step 3, we use Paste operation to get 'AAA'.

Note:

The 

n

 will be in the range [1, 1000].

分析

是否要复制当前的字符串再粘贴，要判断总长度是否是当前的字符串的长度的倍数。如果是倍数关系，则可以复制当前的字符串。复制粘贴当前的字符串需要的次数是i/j，如果这次的次数比之前的次数少，则更新这个次数。

代码

class Solution {
public:
int minSteps(int n) {
if (n == 1) return 0;
int* dp = new int[n+1];
for (int i = 0; i <= n; i++) dp[i] = i;
for (int i = 1; i <= n; i++) {
for (int j = i/2; j > 0; j--) {
if (i % j == 0) {
dp[i] = (dp[j]+i/j) < dp[i] ? (dp[j]+i/j) : dp[i];
break;
}
}
}
int res = dp
;
delete []dp;
return res;
}
};

附

题目地址：LeetCode 650